Re: GCH vs. Axiom of Choice.

From: J.E. (troubled6man_at_yahoo.com)
Date: 12/15/04 

Date: 15 Dec 2004 08:22:36 -0800

KRamsay wrote:
> In article <39d6e584.0412022029.767e52f2@posting.google.com>,
> troubled6man@yahoo.com (J.E.) writes:
> |Ah, I should have been more clear. L is a model (right?),
>
> It's a proper class of sets, the sets satisfying a certain condition.
> (Being "constructible".) Another way to describe it is that it's the
> minimal model of ZF containing all the ordinals.

Is L a set? Obvious L is not in L, but if seems like definitions of say
0# seem to say that it exists if L exists. Is that because the
existence of L is undecideable because otherwise ZF could prove it's
own consistency by proving the existence of L?

[snip Q&A I understood, yeah!]

> | For instance if we had to assume ZFC+GCH
> |to prove it, then it wouldn't mean much IMO.
>
> Heh, it's sort of amusing when to prove that some fact holds true
> when relativized to a model one needs to use the same fact not
> relativized to the model. I can see how that might seem
underwhelming,
> but it's not necessarily trivial or uninteresting. For instance, one
> had to prove that the axioms of ZF hold in L, too, and obviously we
> need to assume some of them in order to prove they hold in L.

I meant that if we prove GCH in L without assuming GCH then it's a bit
like getting GCH for free, and if we have to assume it, then since L is
a standard model (x models x, and xey is true inside the model iff xey
is true outside the model) then I wasn't sure the enterprise was really
generating anything.

> | I assume that ZF can
> |prove only limited things about L, since ZF is incomplete, but if I
> |add GCH to ZF then can ZF+GCH prove something new about L?
>
> I'm guessing it probably can, but I don't know of a good example.
> My guess is that some assumption intermediate between ZF and ZF+V=L
> is sufficient to prove (say) that aleph_1 is the same as aleph_1
> relative to L. Certainly V=L implies that!

Maybe my original question wasn't clear. If you take every sentance X
such that Con(ZF)=>Con(ZF+X) & Con(ZF+~X), then can there be a sentence
S about L (which to more clear "if L is a set", I'll define as a
sentence such that (Ex T) is equivalent to (Ex xeL & T) for every
formulas (Ex T) and (T) that are subformulas of S and such that (Ax T)
is equivalent to (Ax (~xeL)vT) for every formulas (Ax T) and (T) that
are subformulas of S) such that ZF+X|- S and ZF+~X|- ~S. And more
generally, how about any chain of X,Y,Z, ... such that for every subset
A or B of of X,Y,Z,... such that Con(ZF)->Con(ZF+A) and
Con(ZF)->Con(ZF+B) is it possible that ZF+A |- S and ZF+B |- ~S. I'm
trying to figure out if the superconsistent axioms of ZF together proof
the alleged fixed truths of L. FOL is complete, so it might work.

> On the other hand, there aren't any first order properties of L
> that can be proven in ZF+GCH but not in ZF. In fact, for an arbitrary
> sentence X in the language of ZF, the following are equivalent:
>
> (a) ZF |- (V=L->X)
> (b) ZF + V=L |- X
> (c) ZF |- L |= X.

I can read (a) and (b), but (c) and (d) must have some operator order
preference I don't know, or is ZF |- L |= X only to be read as ZF |-
(L|=X), which I don't even know what it means unless L is a sentence,
and even so how one proves a validity based on axioms I don't know so
even then it doesn't make sense to me.

> (d) ZF + V=L |- L |= X
>
> (a)->(b) is a case of the deduction theorem. (b)->(c) holds because
> it's a theorem of ZF that V=L relativized to L is true Given a model
> M of ZF, we can consider constructibility as relativized to M. The
> elements of M satisfying this relativized constructibility form a
> submodel L_M, which is another model of ZF. If we repeat this
process,
> to get L_{L_M}, it's possible to prove that we don't shrink the model
> any; L_{L_M} = L_M. So L_M is a model of V=L, the axiom stating that
> every set is in L. So anything that's a consequence of V=L holds
> relative to L. If X is a theorem of ZF+V=L, then that fact can be
> proven in ZF, and then applied to L.
>
> (c)->(d) is obvious.
>
> To prove (d)->(a), assume first that (a) is false. By Goedel's
> completeness theorem, there would exist a model of ZF+V=L+~X.
> Suppose M is a model of ZF+V=L+~X. Since the L relativized to M
> is the same as M, we get that M |= ~(L|=X). Hence M is also a
> model of ZF+V=L in which L|=X fails.
>
> Since GCH is a consequence of V=L,
>
> (c') ZF + GCH |- L|= X
>
> is also equivalent.
>
> | How about
> |ZF+(~GCH), can that also prove things about L?
>
> Since ~GCH is so weak, I don't think there are many theorems about
> L in ZF+(~GCH) that aren't also theorems in ZF. We can deduce from
> ~GCH the existence of sets not in L, of course.

I'm trying to find out if new relativity consistent axioms added to ZF
prove things about L consistently and whether taken as a whole they do
so completely.

> | Do the truths of L
> |vary depending on what other axioms L is defined with? I'm trying
to
> |be more clear, but I apologize in advance if I failed again.
>
> How would whether something is true of L depend on axioms?

I don't even get the definition of L, so to me it's all provisional.

> Of course, what can be proven to be true of L from a set of axioms
> does depend on the axioms! I'm not familiar with the details, but I
> gather that some large cardinal axioms (I think the existence of a
> measurable cardinal is strong enough) imply a relatively detailed
> account of the structure of L.

But are there other axioms that imply details that contradict those
details?

> Note that the existence of a
> measurable cardinal is inconsistent with V=L.

Yes, of course. I'm just using them to increase the deductive power of
a system to prove the alleged fixed truths of L. Axioms need not be
true, as long as I'm only using them in a way to prove truths of L.

> This real 0# that is believed to be outside of L encodes some of the
> structure of L (if it exists).
Is it L v 0# whose existence is debated?

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